What is the integral of e^{-θ}cos(9θ) ?

The integral of e^{-θ}cos(9θ) is (9e^{-θ}sin(9θ))/(82)-(e^{-θ}cos(9θ))/(82)+C

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integral of e^{-θ}cos(9θ)

\int e^{- θ} cos (9 θ) d θ

\frac{9 e ^{- θ} sin ( 9 θ )}{82} - \frac{e ^{- θ} cos ( 9 θ )}{82} + C

Solution steps

\int e^{- θ} cos (9 θ) d θ

Apply Integration By Parts

= \frac{1}{9} e^{- θ} sin (9 θ) - \int - \frac{1}{9} e^{- θ} sin (9 θ) d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} e^{- θ} sin (9 θ) - (- \frac{1}{9} \cdot \int e^{- θ} sin (9 θ) d θ)

Apply Integration By Parts

= \frac{1}{9} e^{- θ} sin (9 θ) - (- \frac{1}{9} (- \frac{1}{9} e^{- θ} cos (9 θ) - \int \frac{1}{9} e^{- θ} cos (9 θ) d θ))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} e^{- θ} sin (9 θ) - (- \frac{1}{9} (- \frac{1}{9} e^{- θ} cos (9 θ) - \frac{1}{9} \cdot \int e^{- θ} cos (9 θ) d θ))

Therefore

\int e^{- θ} cos (9 θ) d θ = \frac{1}{9} e^{- θ} sin (9 θ) - (- \frac{1}{9} (- \frac{1}{9} e^{- θ} cos (9 θ) - \frac{1}{9} \cdot \int e^{- θ} cos (9 θ) d θ))

Isolate

\int e^{- θ} cos (9 θ) d θ

= \frac{9 e ^{- θ} sin ( 9 θ )}{82} - \frac{e ^{- θ} cos ( 9 θ )}{82}

Add a constant to the solution

= \frac{9 e ^{- θ} sin ( 9 θ )}{82} - \frac{e ^{- θ} cos ( 9 θ )}{82} + C

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What is the integral of e^{-θ}cos(9θ) ?
The integral of e^{-θ}cos(9θ) is (9e^{-θ}sin(9θ))/(82)-(e^{-θ}cos(9θ))/(82)+C